When 100 volt DC source is applied across a coil, a current of 1 A flows through it. When 100 V AC source of 50 Hz is applied to the same coil, only 0.5 A current flows. Calculate the inductance of the coil.

In case of a coil L-R circuit

$I=\frac{V}{Z}, Z=\sqrt{R^{2}+(\omega L)^{2}}$

So when dc is applied, $\omega=0, \mathrm{Z}=\mathrm{R}$

And hence $\mathrm{R}=100 \Omega$

And when ac of 50 Hz is applied

$\begin{array}{l} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}}, \mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{100}{0.5} 200 \Omega \\ \text { But } \mathrm{Z}=\sqrt{\mathrm{R}^{2}+(\omega \mathrm{L})^{2}} \\ 2 \pi \mathrm{fL}^{2}=200^{2}-100^{2}=3 \times 10^{-4} \\ \qquad \mathrm{~L}=\frac{\sqrt{3} \times 10^{2}}{2 \pi \times 50}=\frac{\sqrt{3}}{\pi} \mathrm{H}=0.55 \mathrm{H} \end{array}$

Alternative: To calculate the inductance of the coil, we can use the information provided for both the DC and AC cases. ### Step 1: Determine the Resistance When a DC source is applied: $V = I \times R$ $100 \, \text{V} = 1 \, \text{A} \times R$ $R = 100 \, \Omega$ ### Step 2: Determine the Impedance When an AC source is applied: $V = I \times Z$ $100 \, \text{V} = 0.5 \, \text{A} \times Z$ $Z = \frac{100 \, \text{V}}{0.5 \, \text{A}} = 200 \, \Omega$ ### Step 3: Calculate the Reactance The impedance $Z$ is related to resistance $R$ and inductive reactance $X_L$ by: $Z = \sqrt{R^2 + X_L^2}$ We already know $Z = 200 \, \Omega$ and $R = 100 \, \Omega$: $200 = \sqrt{100^2 + X_L^2}$ $200 = \sqrt{10000 + X_L^2}$ $200^2 = 10000 + X_L^2$ $40000 = 10000 + X_L^2$ $X_L^2 = 30000$ $X_L = \sqrt{30000}$ $X_L \approx 173.2 \, \Omega$ ### Step 4: Calculate the Inductance The inductive reactance $X_L$ is related to the inductance $L$ and the frequency $f$ by: $X_L = 2\pi f L$ Given $f = 50 \, \text{Hz}$ and $X_L \approx 173.2 \, \Omega$: $173.2 = 2\pi \times 50 \times L$ $L = \frac{173.2}{2\pi \times 50}$ $L = \frac{173.2}{314.16}$ $L \approx 0.552 \, \text{H}$ Therefore, the inductance of the coil is approximately $0.552 \, \text{H}$.